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(C) The magnetic field $B$ at a distance $r$ from an infinitely long wire is given by $B = \frac{\mu_0 i}{2 \pi r}$.As the loop moves with velocity $v

Vedclass · Accepted Answer

$\frac{\mu_0 i l b v}{2 \pi x(l+x)}$

Vedclass · Answer

(C) The magnetic field $B$ at a distance $r$ from an infinitely long wire is given by $B = \frac{\mu_0 i}{2 \pi r}$.As the loop moves with velocity $v$,the motional emf is induced in the two vertical sides of length $b$.For the side at distance $x$,the velocity is perpendicular to the magnetic field,so the induced emf is $e_1 = B_1 v b = \left(\frac{\mu_0 i}{2 \pi x}\right) v b$.For the side at distance $(x+l)$,the induced emf is $e_2 = B_2 v b = \left(\frac{\mu_0 i}{2 \pi (x+l)}\right) v b$.Since the loop is moving away,the emf induced in these two sides opposes each other in the loop circuit.The net magnitude of the emf is $e = |e_1 - e_2| = \frac{\mu_0 i v b}{2 \pi} \left( \frac{1}{x} - \frac{1}{x+l} \right)$.Simplifying the expression: $e = \frac{\mu_0 i v b}{2 \pi} \left( \frac{x+l-x}{x(x+l)} \right) = \frac{\mu_0 i l b v}{2 \pi x(x+l)}$.

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